Stokes s Theorem 17.2
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1 Stokes s Theorem December 213 Stokes s Theorem is the generalization of Green s Theorem to surfaces not just flat surfaces (regions in R 2 ). Relate a double integral over a surface with a line integral over the surface s boundary Need to orient the boundary Involves the curl of a vector field Orientation of Surface Boundary. Say S is a surface: boundary S, finite union of simple closed curves S oriented with normals n Then S oriented by the right-hand rule: point your right thumb along n with your palm facing INTO S then your fingers point in the direction to orient S Or: If walking along S with head pointing up (n) then S is on left hand side.
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3 Curl of Vector Field. Given vector field F = F 1, F 2, F 3, the curl of F is curl(f) = F = det î ĵ ˆk x y z F 1 F 2 F 3 Example: For F = y 3, y + z, z x, F = det î ĵ ˆk x y z y 3 y + z z x (x z) (y + z) =,... y z = 1, 1, 3y 2.
4 Example, Curl. For F(x, y, z) = 2x + 7y + 5z, x 3 + y 3 + z 3, cos(y + z), î ĵ ˆk curl(f) = F = det x y z 2x+7y+5z x 3 +y 3 +z 3 cos(y+z) = ( sin(y + z) 3z 2 )î ( 5)ĵ + (3x 2 7)ˆk = sin(y + z) 3z 2, 5, 3x 2 7. Worksheet. Worksheet #1 4 Compute the curls of the vector fields in problems #1 4. Reminder: an integral like xz dx + yz dy + xy dz means the vector field F = xz, yz, xy. C Physical Interpretation of Curl. If F is the velocity field of a fluid flow and a small sphere is located at P and allowed to rotate but not to move, then the fluid will spin the sphere according to curl(f) P. Point your right thumb along curl(f) P. Then your fingers will curl in the direction that the sphere spins. The magnitude of curl(f) P indicates the speed of the spin.
5 Stokes s Theorem. Stokes s Theorem S F ds = curl(f) ds The circulation of F around the boundary S is equal to the flux of curl(f) through S. Green s theorem is when S is flat (S R 2 ) with normal n =,, 1. Then, curl(f) n = curl z (F) = F 2x F 1y. S Example, Stokes s Theorem. Example: For F(x, y, z) = 2x y, y + 7z, 5x + z, S the upper unit hemisphere x 2 + y 2 + z 2 = 1, z, oriented with upward-pointing normal, verify Stokes s Theorem. (That is, compute both sides of the equation in Stokes s Theorem and verify that they give the same value.) Example continued: Line Integral. F(x, y, z) = 2x y, y + 7z, 5x + z Line Integral: Compute the line integral about the boundary, which is Then F ds = S (x, y, z) = c(θ) = (cos θ, sin θ, ), = = 2π 2π 2π = = π. θ 2π 2 cos θ sin θ, sin θ, 5 cos θ sin θ, cos θ, dθ cos θ sin θ + sin 2 θ dθ 2π cos θ( sin θ) dθ + 1 cos 2 θ dθ 2
6 Example continued: Surface Integral of Curl. The curl is F = det F(x, y, z) = 2x y, y + 7z, 5x + z î ĵ ˆk x y z = 7, 5, 1 2x y y + 7z 5x + z S is parametrized by (x, y, z) = (cos θ sin φ, sin θ sin φ, cos φ), θ 2π, φ π/2, so T θ = sin θ sin φ, cos θ sin φ,, T φ = cos θ cos φ, sin θ cos φ, sin φ, T θ T φ = sin φ cos θ sin φ, sin θ sin φ, cos φ. This is downward-pointing so let s use n = T θ T φ = sin φ cos θ sin φ, sin θ sin φ, cos φ. F = 7, 5, 1, Then n = sin φ cos θ sin φ, sin θ sin φ, cos φ. curl(f) n = sin φ ( 7 cos θ sin φ 5 sin θ sin φ + cos φ), so curl(f) ds = π/2 2π S π/2 = 2π = 2π 1 2 sin2 φ = π. sin φ ( ) dθ dφ sin φ cos φ dφ π/2 Conservative Vector Fields and Curl. Notice: F = det î ĵ ˆk x y z = F 3y F 2z, F 1z F 3x, F 2x F 1y F 1 F 2 F 3 so F conservative path-independent exact = closed curl(f) = One of the upshots: If F is a gradient vector field, F = φ, then F is zero: ( φ) = or curl(grad φ) =
7 Closed Surfaces. Notice: If S is a closed surface, meaning S is empty, e.g., sphere, box, torus, any surface that completely encloses a region of space, then curl(f) ds =! S Because F ds = automatically. S Example, Closed Surface. Example: Compute the flux of A(x, y, z) = 2x, 7y, 5z across the torus T, (r 16) 2 + z 2 = 16 (see picture on board) Answer: It just so happens that A = curl(f), where So, T F = 5yz,, 2xy. A ds = curl(f) ds = F ds =. T T The flux of A through T is zero. Surfaces with Same Boundary. Notice: If S 1 and S 2 are two surfaces with the same boundary, and if A happens to be a curl vector field (A = curl(f)) then the fluxes of A through S 1 and S 2 are equal: S 1 A ds = S 2 A ds. This is similar to: If C 1 and C 2 are two curves with the same endpoints, and if F happens to be a gradient vector field (F = grad(φ)) then F is path-independent: C 1 F ds = C 2 F ds.
8 Example, Surfaces with Same Boundary. Example: Let S be the upper half of the ellipsoid x 2 + y 2 + 4z 2 = 1, z, oriented with upward-pointing normal. Let A = 2 xz, 3 yz, 4 + z 2. What is the flux of A through S? Answer: First, notice that S has the same boundary as the unit disk D in the xy-plane. Second, it happens that A is a curl vector field: A = F for F = 1 2 yz 2, 4x xz 2, 3x + 2y Therefore, the flux of A across S is the same as the flux of A across the disk D. Example continued. To compute the flux of A across the disk D, D has the normal vector n = ˆk =,, 1 A n = 4 + z 2 = 4 because z =, since D contained in xy-plane Therefore flux = D A ds = 4 ds = 4 Area(D) = 4π. D Worksheet. Worksheet #1 4
9 Clicker Question: The pictures below show top views of three vector fields, all of which have no z component. Which one has the curl pointing in the positive ˆk direction at the origin (the center of each picture)? A. the one on the left B. the one in the middle receiver channel: 41 C. the one on the right D. none of them session ID: bsumath275 Clicker Question: The vector field F has curl F = 3, 4, 2. What is the magnitude of the circulation of F around the perimeter of the square with corners at coordinates (1, 2, 3), (4, 2, 3), (4, 2, 6), and (1, 2, 6)? A. B. 18 C. 27 D. 36 E. 81 F. None of the above receiver channel: 41 session ID: bsumath275
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